As stated, the answer is trivially no, since as you observe, the condition forces $p$ to also vanish on the coordinate hyperplanes, which are not contained in the positive orthant. If you modify the question to say that $p(x)>0$ iff $x$ is in the open positive orthant, the answer is still no (for $n\geq 2$). As you observe, we can write $p(x)=x_1\dots x_nq(x)$ for some $q$. But then $q$ must change sign when you change between any two adjacent orthants that are not the positive orthant (since $x_1\dots x_n$ changes sign but $p$ does not), so $q$ is also divisible by $x_1\dots x_n$. We can thus write $p(x)=x_1^2\dots x_n^2r(x)$ where $r$ is also positive only in the positive orthant, and so by induction on degree no such $p$ can exist.

I'm not sure what kind of "approximation" you're looking for, but here's a fairly strong sense in which no approximation can exist with nice uniformity properties as $x$ gets large. Let $p$ be any nonconstant polynomial in $n$ variables and consider the single variable polynomials $q_x(t)=p(tx)$. For almost every $x$ (i.e., for $x$ not lying in a proper subvariety), $q_x$ will have the same degree as $p$. In particular, if $p$ has even degree then for almost every $x$, $p(tx)$ will have the same sign as $t\to\infty$ and $t\to-\infty$, and if $p$ has odd degree $p(tx)$ will almost always have opposite sign as $t\to \infty$ and $t\to -\infty$. On the other hand, a function that is positive in the positive orthant and negative in all the other orthants does not behave like either an even or an odd degree polynomial in this respect. In particular, any nonconstant polynomial "approximation" must be large and have the wrong sign on (asympotically) at least $1/2^n$ of the volume of large balls around the origin.